Subjective Questions
Discrete Mathematics: Combinatorics and Probability
Chapter 1: Introduction to Discrete Mathematics
Discrete Mathematics: Combinatorics and Probability
Introduction:
Welcome to the fascinating world of Discrete Mathematics! In this chapter, we will delve into the realm of Combinatorics and Probability, two fundamental areas of study in mathematics. Grade 10 is an important year for students as they lay the groundwork for their future mathematical endeavors. Understanding the principles of combinatorics and probability is crucial for success in higher level math courses and real-life applications. In this chapter, we will explore the basics of these concepts and provide you with a comprehensive understanding of the subject matter.
Section 1: Combinatorics
1.1 Permutations:
Permutations are an essential concept in combinatorics. They involve the arrangement of objects in a specific order. For example, if we have three letters A, B, and C, the number of ways we can arrange them is given by the formula nPr = n! / (n-r)!, where n is the total number of objects and r is the number of objects to be arranged. Let\’s illustrate this with a simple example.
Example 1: Consider the letters A, B, and C. How many different arrangements can be made using all three letters?
Solution: Using the formula nPr = n! / (n-r)!, we have 3P3 = 3! / (3-3)! = 3! / 0! = 3! / 1 = 3.
1.2 Combinations:
Combinations, on the other hand, involve the selection of objects without considering their order. For example, if we have three letters A, B, and C, the number of ways we can select two letters is given by the formula nCr = n! / (r!(n-r)!), where n is the total number of objects and r is the number of objects to be selected. Let\’s take a look at a medium-level example.
Example 2: Consider the letters A, B, C, D, and E. How many different combinations of two letters can be formed?
Solution: Using the formula nCr = n! / (r!(n-r)!), we have 5C2 = 5! / (2!(5-2)!) = 5! / (2!3!) = (5*4) / (2*1) = 10.
1.3 Binomial Theorem:
The Binomial Theorem is a powerful tool in combinatorics that allows us to expand expressions of the form (a + b)^n. It provides a systematic way to determine the coefficients of each term in the expansion. Let\’s explore a complex example to understand this concept better.
Example 3: Expand (x + y)^4.
Solution: Using the Binomial Theorem, we can expand this expression as follows:
(x + y)^4 = C(4,0)x^4y^0 + C(4,1)x^3y^1 + C(4,2)x^2y^2 + C(4,3)x^1y^3 + C(4,4)x^0y^4
= 1x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 1y^4
= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.
Section 2: Probability
2.1 Fundamentals of Probability:
Probability is the branch of mathematics that deals with the likelihood of events occurring. It is a crucial concept in various fields, including statistics, economics, and computer science. To understand probability, we need to understand the basic principles and terminology associated with it. Let\’s explore a simple example to get started.
Example 4: What is the probability of rolling a fair six-sided die and getting an even number?
Solution: Since there are three even numbers (2, 4, and 6) out of a total of six possible outcomes, the probability is 3/6 or 1/2.
2.2 Conditional Probability:
Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), where A and B are events. To calculate conditional probability, we use the formula P(A|B) = P(A and B) / P(B). Let\’s examine a medium-level example.
Example 5: A bag contains 4 red balls and 6 blue balls. If two balls are drawn without replacement, what is the probability of getting two red balls?
Solution: The probability of drawing the first red ball is 4/10. Since one ball has been removed, there are now 9 balls left in the bag, with 3 red balls remaining. Therefore, the probability of drawing a second red ball is 3/9. The probability of both events occurring is (4/10) * (3/9) = 2/15.
2.3 Bayes\’ Theorem:
Bayes\’ Theorem is a fundamental concept in probability theory that allows us to update our beliefs about the likelihood of an event occurring based on new evidence. It is particularly useful in fields such as medicine, where diagnostic tests are involved. Let\’s take a look at a complex example to understand this concept better.
Example 6: A certain disease affects 1% of the population. A test for the disease has a 95% accuracy rate. If a person tests positive, what is the probability that they have the disease?
Solution: Let A be the event of having the disease and B be the event of testing positive. We are looking for P(A|B). Using Bayes\’ Theorem, we have P(A|B) = (P(B|A) * P(A)) / P(B), where P(B|A) is the accuracy rate of the test, P(A) is the probability of having the disease, and P(B) is the probability of testing positive. Plugging in the values, we have P(A|B) = (0.95 * 0.01) / (0.95 * 0.01 + 0.05 * 0.99) ≈ 0.161.
Chapter 2: Subjective Questions and Detailed Reference Answers
Question 1: In how many ways can the letters of the word \”MATHEMATICS\” be arranged?
Solution: The word \”MATHEMATICS\” consists of 11 letters, including 2 M\’s, 2 A\’s, and 2 T\’s. Using the formula for permutations with repetition, the total number of arrangements is 11! / (2!2!2!) = 4989600.
Question 2: A committee of 5 people is to be chosen from a group of 10 men and 8 women. In how many ways can this be done if the committee must consist of at least 2 men and 2 women?
Solution: We can approach this problem by considering two cases. In the first case, we select 2 men and 3 women from the respective groups. This can be done in C(10,2) * C(8,3) = 45 * 56 = 2520 ways. In the second case, we select 3 men and 2 women, which can be done in C(10,3) * C(8,2) = 120 * 28 = 3360 ways. Therefore, the total number of ways to form the committee is 2520 + 3360 = 5880.
Question 3: A bag contains 5 red balls and 7 blue balls. If 3 balls are drawn at random without replacement, what is the probability that all three balls are blue?
Solution: The probability of drawing the first blue ball is 7/12. Since one ball has been removed, there are now 11 balls left in the bag, with 6 blue balls remaining. Therefore, the probability of drawing a second blue ball is 6/11. Similarly, the probability of drawing a third blue ball is 5/10. The probability of all three events occurring is (7/12) * (6/11) * (5/10) = 7/44.
Question 4: A fair coin is tossed 4 times. What is the probability of getting at least 3 heads?
Solution: The probability of getting exactly 3 heads is given by the formula C(4,3) * (1/2)^3 * (1/2)^1 = 4/16 = 1/4. The probability of getting exactly 4 heads is (1/2)^4 = 1/16. Therefore, the probability of getting at least 3 heads is 1/4 + 1/16 = 5/16.
Question 5: In a deck of 52 cards, what is the probability of drawing a red card followed by a black card without replacement?
Solution: The probability of drawing a red card from a deck of 52 cards is 26/52 = 1/2. Since one card has been removed, there are now 51 cards left in the deck, with 26 black cards remaining. Therefore, the probability of drawing a black card is 26/51. The probability of both events occurring is (1/2) * (26/51) = 26/102.
Question 6: A box contains 6 red balls and 4 blue balls. If 2 balls are drawn at random with replacement, what is the probability that both balls are red?
Solution: The probability of drawing a red ball on the first draw is 6/10. Since the ball is replaced, the probability of drawing a red ball on the second draw is also 6/10. The probability of both events occurring is (6/10) * (6/10) = 36/100 = 9/25.
Question 7: A box contains 5 red balls, 3 blue balls, and 2 green balls. If 3 balls are drawn at random without replacement, what is the probability that at least 2 balls are red?
Solution: We can approach this problem by considering two cases. In the first case, we select 2 red balls and 1 non-red ball from the respective groups. This can be done in C(5,2) * C(5,1) = 10 * 5 = 50 ways. In the second case, we select all 3 red balls, which can be done in C(5,3) = 10 ways. Therefore, the total number of ways to draw at least 2 red balls is 50 + 10 = 60. The total number of ways to draw 3 balls without replacement is C(10,3) = 120. Therefore, the probability is 60/120 = 1/2.
Question 8: A fair six-sided die is rolled twice. What is the probability of getting a sum of 7?
Solution: There are 6 possible outcomes for each roll of the die. To determine the total number of outcomes, we multiply the number of outcomes for each roll together, giving us 6 * 6 = 36. Out of these 36 outcomes, there are 6 ways to get a sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). Therefore, the probability of getting a sum of 7 is 6/36 = 1/6.
Question 9: A bag contains 8 red balls and 5 green balls. If 2 balls are drawn at random with replacement, what is the probability of getting a red ball followed by a green ball?
Solution: The probability of drawing a red ball from a bag of 13 balls is 8/13. Since the ball is replaced, the probability of drawing a green ball is 5/13. The probability of both events occurring is (8/13) * (5/13) = 40/169.
Question 10: A bag contains 3 red balls and 4 blue balls. If 2 balls are drawn at random without replacement, what is the probability of getting 2 balls of the same color?
Solution: We can approach this problem by considering two cases. In the first case, we select 2 red balls from the bag, which can be done in C(3,2) = 3 ways. In the second case, we select 2 blue balls, which can be done in C(4,2) = 6 ways. Therefore, the total number of ways to draw 2 balls of the same color is 3 + 6 = 9. The total number of ways to draw 2 balls without replacement is C(7,2) = 21. Therefore, the probability is 9/21 = 3/7.
Question 11: A fair coin is flipped 5 times. What is the probability of getting exactly 3 tails?
Solution: The probability of getting exactly 3 tails is given by the formula C(5,3) * (1/2)^3 * (1/2)^2 = 10/32 = 5/16.
Question 12: A box contains 6 red balls, 5 blue balls, and 4 green balls. If 3 balls are drawn at random with replacement, what is the probability that all three balls are blue?
Solution: The probability of drawing a blue ball on each draw is 5/15. Since the balls are replaced after each draw, the probability remains constant. Therefore, the probability of all three events occurring is (5/15) * (5/15) * (5/15) = 125/3375 = 1/27.
Question 13: A box contains 7 red balls and 9 blue balls. If 2 balls are drawn at random without replacement, what is the probability of getting a red ball followed by a blue ball?
Solution: The probability of drawing a red ball from a box of 16 balls is 7/16. Since one ball has been removed, there are now 15 balls left in the box, with 9 blue balls remaining. Therefore, the probability of drawing a blue ball is 9/15. The probability of both events occurring is (7/16) * (9/15) = 63/240 = 21/80.
Question 14: A fair six-sided die is rolled three times. What is the probability of getting a sum of 10 or more?
Solution: There are 6 possible outcomes for each roll of the die. To determine the total number of outcomes, we multiply the number of outcomes for each roll together, giving us 6 * 6 * 6 = 216. Out of these 216 outcomes, there are 27 ways to get a sum of 10 or more: (4, 4, 2), (4, 3, 3), (3, 4, 3), (3, 3, 4), (4, 2, 4), (2, 4, 4), (5, 5, 2), (5, 2, 5), (2, 5, 5), (5, 4, 1), (5, 1, 4), (4, 5, 1), (4, 1, 5), (1, 5, 4), (1, 4, 5), (6, 4, 0), (6, 0, 4), (4, 6, 0), (4, 0, 6), (0, 6, 4), (0, 4, 6), (5, 5, 0), (5, 0, 5), (0, 5, 5), (6, 3, 1), (6, 1, 3), (3, 6, 1), (3, 1, 6), (1, 6, 3). Therefore, the probability of getting a sum of 10 or more is 27/216 = 1/8.
Question 15: A bag contains 4 red balls, 3 blue balls, and 2 green balls. If 3 balls are drawn at random without replacement, what is the probability that at least 2 balls are red?
Solution: We can approach this problem by considering three cases. In the first case, we select 2 red balls and 1 non-red ball from the respective groups. This can be done in C(4,2) * C(5,1) = 6 * 5 = 30 ways. In the second case, we select all 3 red balls, which can be done in C(4,3) = 4 ways. In the third case, we select 2 red balls and 1 green ball, which can be done in C(4,2) * C(2,1) = 6 * 2 = 12 ways. Therefore, the total number of ways to draw at least 2 red balls is 30 + 4 + 12 = 46. The total number of ways to draw 3 balls without replacement is C(9,3) = 84. Therefore, the probability is 46/84 = 23/42.
In this chapter, we have covered the basics of combinatorics and probability. We explored permutations, combinations, and the Binomial Theorem in combinatorics, while also delving into the fundamentals of probability, conditional probability, and Bayes\’ Theorem. Additionally, we provided detailed solutions to 15 subjective questions commonly asked in Grade 10 examinations, covering a wide range of concepts and difficulty levels. By thoroughly understanding the principles and techniques discussed in this chapter, you will be well-prepared to tackle combinatorics and probability problems and excel in your mathematical journey. Remember to practice regularly and apply these concepts to real-life situations to solidify your understanding. Good luck!